已知函数f(x)=sin^2wx+√3sinwxcoswx,x属于R,又f（a）=-1/2,f（b）=1/2,若|a-b|的最小值为3π/4,则正数w的值为?
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f(x)=sin²wx+根号下3sinwxcosx
=(1-cos2wx)/2+根号下3sin2wx/2
=sin(2wx-π/6)+1/2
f(a)=sin(2wa-π/6)+1/2=-1/2
sin(2wa-π/6)=-1
2wa-π/6=2kπ-π/2
a=(kπ-π/6)/w
f(b)=sin(2wb-π/6)+1/2=1/2
sin(2wb-π/6)=0
2wb-π/6=2kπ
b=(kπ+π/12)/w
|a-b|=的最小值为3π/4
|π/4|/w=3π/4
w=1/3